Answer to Question #246940 in Differential Equations for Jac

Question

Solve the differential equation by Bernoulli equation. Show complete solution.

dy/dx + 1/3 y = 1/3 (1 + 3x) y⁴

Answer 1

$$y'+\dfrac{1}{3}y=\dfrac{1}{3}(1+3x)y^4$$

$$v=y^{1-n}=y^{1-4}=y^{-3}$$

$$v'=-3y^{-4}y'$$

$$-\dfrac{1}{3}v'+\dfrac{1}{3}v=\dfrac{1}{3}(1+3x)$$

$$v'-v=-1-3x$$

Integrating factor

$$\mu(x)=e^{-x}$$

$$e^{-x}v'-e^{-x}v=-e^{-x}(1+3x)$$

$$d(e^{-x}v)=-e^{-x}(1+3x)dx$$

Integrate

$$\int d(e^{-x}v)=-e^{-x}(1+3x)dx$$

$$\int xe^{-x}dx$$

$$\int udv=uv-\int vdu$$

$$u=x, du=dx$$

$$dv=e^{-x}dx, v=\int e^{-x}dx=-e^{-x}$$

$$\int xe^{-x}dx=-xe^{-x}+\int e^{-x}dx$$

$$=-xe^{-x}-e^{-x}+C_1$$

$$e^{-x}v=e^{-x}+3xe^{-x}+3e^{-x}+C$$

$$v=4+3x+Ce^x$$

$$y=\dfrac{1}{\sqrt[3]{4+3x+Ce^x}}$$